Finding x in Logarithmic Equation

[hola1]

Hi, sorry if it's not in the right subforum. idk how to solve x:
http://puu.sh/2Lbb1.png
The answer is x = -1.70951.
how do we get there? please explain everystep. thanks :3

****someone made it this far, idk if it is the correct path:
log2 (2^(x-1)+3^(x+1)) = 2x - log2 (3^x)
log2 (2^(x-1)+3^(x+1)) + log2 (3^x) = 2x
because of the rule log(m) + log(n) = log(mn),
log2 ((2^(x-1)+3^(x+1))*(3^x) = 2x
log2 ((2^(x-1)+3^(x+1))*(3^x) = 2x
log ((2^(x-1)+3^(x+1))*(3^x) / log 2 = 2x
log ((2^(x-1)+3^(x+1))*(3^x) = 2x * log 2
log ((2^(x-1)+3^(x+1))*(3^x) = log 2^(2x)
equate the logs
(2^(x-1) + 3^(x+1))*(3^x) = 2^(2x)
2^(x-1) * 3^x + 3^(2x+1) = 2^(2x)
3^(2x+1) = 2^(2x) - 2^(x-1) * 3^x

 

[tkhunny]

Re: logarithm

Seems like kind of a struggle, but you are getting good practice playing with the logarithms.

I might do this:

\log_{2}\left(2^{x-1}+3^{x+1}\right) = 2x - \log_{2}\left(3^{x}\right) = \log_{2}\left(2^{2x}\right)- \log_{2}\left(3^{x}\right) = \log_{2}\left(\dfrac{2^{2x}}{3^{x}}\right)

This leads a little more quickly to a version with no logs which may not be as useful as you think.

2^{x-1} + 3^{x+1} = 2^{2x}\cdot 3^{-x} = \left(\dfrac{4}{3}\right)^{x}

There is no way to solve that, so you are really left with numerical methods, which probably causes you to reintroduce the logarithms.

Can you take it from there?

I get x = -1.70951129135145, which certainly agrees with your given solution.